GATE_PROJ
Top 16 Positive Activations
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8 K (25°C) yields ~168.5 m/s. Therefore, since the
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and rounded to three sig figs is 169 m/s. Thus, present answer boxed as
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compute", so it's expected to compute.\n\nAlternatively, perhaps check if there is an established average speed for
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precision needed.\n\nAlternatively, maybe the original question assumes a specific temperature but forgot to mention. If the user
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and calculator computations, perhaps the answer expects 170 m/s as an approximate answer. But some
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98 K (25°C) yields ~168.5 m/s. Therefore, since
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. Since the user hasn't specified temperature, perhaps there's a standard temperature to assume. Let me check
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different.\n\nBut given that the user hasn't specified the temperature, this seems ambiguous. However, in many
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is rounded to three significant figures, giving \(169 \, \text{m/s}\).\n\n
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around 298 K. Maybe. Alternatively, as room temperature sometimes is taken as 30
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's expected to compute.\n\nAlternatively, perhaps check if there is an established average speed for radon gas at
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, given that no temperature is given, perhaps they just want the formula, but I don't think so
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, if forced to choose, answer approximately 170 m/s. But given compute precisely, it
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m/s. But if the question hasn't specified the temperature, this can vary.\n\nAlternatively, given that
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we need to assume a standard temperature.\n\nAlternatively, maybe they expect us to recall the molar mass of
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the absence of a specified temperature, perhaps we need to answer that the temperature is required, but the problem
Top 16 Negative Activations
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Let's see.\n\n168^2 = (170 - 2)^2 =
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0×30 =7200, 7440-7200=
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,030,000, since 5025*1000=
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24=14,328 and 597*0.8004
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0. Let's see: 240×30 =7200, 7
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by 240. Let's see: 240×30 =720
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0 by 5025:\n\nFirst, 5025 * 1200
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6,000,000 and 2000*15=30
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. 891 mod125: 125*7=875.
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125-120=5; 125≡5 mod8.\n\nSimilarly
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=1000 mod125. 1000 divided by125 is
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5,920,250. 5,940,150 -
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40×30 =7200, 7440-7200
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1 mod125: 125*7=875. 891
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's see: 240×30 =7200, 7440
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=30,150. Then, 30,150 /502
UP_PROJ
Top 16 Positive Activations
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but their exponents are 1, which is odd. Actually, wait nvm, the exponents
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). The altitude is h, the horizontal distance from G to EM. So h is the length we need
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7 terms) is-1 mod1000. So multiplying 891 by (-1
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:4,7:2, which are all even? Wait, 18,8,4
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, so 0.88% is already two significant figures. Alternatively, if written as 0
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= sqrt(2) * ||x - y|| / sqrt{1 - a^, but this
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care. So GEM is an isosceles triangle with base EM, so EM is from (
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128\), since each pair is counted twice. But actually, since we are considering ordered pairs
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exponent of 7 is 2, which is even. Wait a second. Then primes 11
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- y) : x ∈ S } are orthonormal. Hence, they can be considered as an
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the vectors from the centroid to the vertices are orthonormal. Wait, not exactly, but in the
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each unordered pair {a, b} is counted twice in the 256 ordered pairs.\n\nBut
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k -1= -1 mod1000 forall k>=3. Therefore, product becomes9
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ormal system. That means each vector has length 1 and any two distinct vectors are orthogonal.\n\nFirst,
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256\), but each pair is counted twice except when \(a = b\). But since
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18 for exponent of 2. Which is even. 3: 8, even.
Top 16 Negative Activations
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time.\n\nSo perhaps the answer is not straightforward, hence why reference to the P/poly theorem. The
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in circles here. Given the time constraints, maybe for problem 3, since it's known in complexity
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only countable ones.\n\nBut the question is probably supposed to allow the System S to be mapped to or
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P/poly.\n\nTherefore, for the first problem, combining multiple sparse sets into one via tagging. For
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the subject.\n\nAlternatively, but, finally, the answer is asserting that such a y exists because he’s
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problem asks to show it, not just cite, perhaps construction:\n\nGiven L ∈ P_angel, then
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Given the time constraints, maybe for problem 3, since it's known in complexity theory that P/p
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'm going in circles here. Given the time constraints, maybe for problem 3, since it's known
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going in circles here. Given the time constraints, maybe for problem 3, since it's known in
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.\n\nSo perhaps the answer is not straightforward, hence why reference to the P/poly theorem. The connection
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such a y exists. Thus, finally, the answer is "\boxed{y \text{ exists as
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circles here. Given the time constraints, maybe for problem 3, since it's known in complexity theory
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polynomial time.\n\nSo perhaps the answer is not straightforward, hence why reference to the P/poly theorem.
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that, after struggling through a number of different avenues, I'll try to formalize the key steps.\n\n
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uses the separability and completeness.\n\nBut maybe an innovative approach is to use the idea of completing the system
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the definition of P/poly.\n\nTherefore, for the first problem, combining multiple sparse sets into one via
DOWN_PROJ
Top 16 Positive Activations
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sqrt(0.000064) is 0.008, and sqrt
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=30,150. Then, 30,150 /502
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0, since 5025*1000=5,025,0
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14,328 and 597*0.8004≈ 5
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=19,900. Then, 19,900 /497
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,000 and 2000*15=30,000 and
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function f(x) = 2⟨x, y⟩ - ⟨x, x⟩ is constant on
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.\n\nHence, the solution is x≡109 mod1000. Therefore, P
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have ⟨x, x'⟩ - ⟨x, y⟩ - ⟨x', y⟩ + ⟨
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6,000,000 and 2000*15=30
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3*66.512:\n\n70*66.512=4,
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*(1000 -1)=891,000 -891=8
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see. I need to solve this problem about a trapezoid with bases differing by 10
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(10, 10), and E is at (10, 0). So the
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0 divided by125 is8, so 1000≡0 mod125
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= ⟨x, x'⟩ - ⟨x, y⟩ - ⟨x', y⟩ + ⟨
Top 16 Negative Activations
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1200. Yep, that's right. So, C_eq is 120
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,0)):\n\nThe slope (m) is (0 -5)/(10 - (10
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8≈168, which is about right. So, the number looks consistent. Therefore, if
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respectively. Then, error terms calculated for each. \n\nAlternatively, perhaps I could use logarithmic differentiation.
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..., up to k=999 (10^999 -1). Therefore, the
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,2) = same 60*20= 1200. Times 2
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0 - (10 - h)) = (-5)/h.\n\nThe equation is y -0 =
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5*4)=2*60*20=2400\n\nk=4:\n\n
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sequences: R GGGG, GGGG R\n\nk = 2: RR GGG,
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, etc., would give the same result as before.\n\nLet me actually compute it this way:\n\nFirst term
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:\n\nThe slope (m) is (0 -5)/(10 - (10 - h))
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E(10,0)):\n\nThe slope (m) is (0 -5)/(10
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formula for percentage error in energy would then be the square root of the sum of the squares of the relative
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10,10).\n\nDistance GM = sqrt[(10 - (10 - h))^2
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00\n\nk=3:\n\nSame as k=2: 2*P(5,3
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109 is odd, which is consistent.\n\nSimilarly, modulus 8 of5 is also odd